2
$\begingroup$

I am asked to show that the series below converges:

$$\sum_{n=1}^{\infty} \left(1 + \frac{(-1)^n-3}{n}\right)^{n^2}$$

I am not very well versed on this subject and I believe I have found a valid proof, but I am not sure.

Side note, for my proof I used what I call Cauchy's criterion [not sure if this is standard terminology]:

Cauchy's criterion: Let $\{x_n\}$ be a succession with $x_n \geq 0$. If $\limsup_{n \to \infty} \sqrt[n]{x_n} < 1$ then $\sum x_n$ converges. If $\limsup_{n \to \infty} \sqrt[n]{x_n} > 1$ then $\sum x_n$ diverges.

Using that criterion, we calculate the $\limsup$ of the general term:

$$\limsup_{n \to \infty} \sqrt[n]{\left(1 + \frac{(-1)^n - 3}{n}\right)^{n^2}} = \limsup_{n \to \infty} \left(1 + \frac{(-1)^n - 3}{n}\right)^n =\\ \lim_{n \to \infty} \left(1 + \frac{1 - 3}{n}\right)^n = \lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n = e^{-2} < 1$$

The only doubt I have is when calculating the $\limsup$ part, because I was only introduced to it today. If I got it right, the $\limsup$ of a succession can be evaluated by taking the limit of the greatest sub-succession of the original succession.

Is my proof correct/did I calculate the $\limsup$ correctly?

Thanks for your time.

  • 3
    It looks fine to me. – 2017-01-24

2 Answers 2

1

Your approach is fine. I'm not sure if you were required or encouraged to apply the root test, but thought it might be useful to present an alternative way forward. To that end we proceed.

Recalling that $\log(1-x)\le -x$ and $e^x\ge \frac12x^2$ for $x>0$, we have

$$\begin{align} \left(1+\frac{(-1)^n-3}{n}\right)^{n^2}&=e^{n^2\log\left(1+\frac{(-1)^n-3}{n}\right)}\\\\ &\le e^{n^2\log\left(1-\frac{2}{n}\right)}\\\\ &\le e^{-2n}\,\,\,\,\,\dots \text{Note} \sqrt[n]{e^{-2n}}=e^{-2}\\\\ &=\frac{1}{e^{2n}}\\\\ &\le \frac{1}{2n^2} \end{align}$$

Inasmuch as $\sum_{n=1}^\infty\frac{1}{2n^2}=\pi^2/12$, the series of positive terms for $n\ge 4$, $\left(1+\frac{(-1)^n-3}{n}\right)^{n^2}$, converges by comparison.

0

Alternate approach: For large $n,$ the terms are positive and bounded above by $(1-2/n)^{n^2} = ((1-2/n)^{n})^n.$ We know $(1-2/n)^n \to 1/e^2<1/e.$ So for large $n,$ $(1-2/n)^n < 1/e.$ Thus the terms are $< (1/e)^n$ for large $n.$ Since $\sum (1/e)^n <\infty,$ the original series converges.